31) Verify that Stokes' Theorem is true for the vector field F(x,y,z) = x² i + y² j + z² k, where S is the part of the paraboloid z 1 - x² - y ² that lies above the xy-plane and S has an upward orientation. 2

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Q: Compute feF. dr where C is the curve of intersection of the paraboloid z=9-x² - y² and the plane z =…

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Q: F(x, y, z) = (x²)Î + (2y³2)ĵ + (32)Â is a vector field. (a) div(F) (b) curl(F) (c) C is the…

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Q: Verify Stokes theorem for the vector field F = 2(-+x)i- z? yj-zyk over the upper half x² +y? +z?…

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Q: A) Consider the vector field F(x, y, z) = (-3yz, 6xz, xy). Find the divergence and curl of F. 0…

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Q: A vector field is given by A(x,y,z)=xyi+2yzj+3zxk. In this multi-part question we will check that…

A: Here, the vector field, A(x,y,z)=xy i + 2yz j + 3zx k. So, the curl A=ijk∂∂x∂∂y∂∂zxy2yz3zx ⇒curl…

Q: Let vector field F =(x,y,z²) across the surface S which is an upper hemisphere x²+ y² + z² = 4 . 1)…

A: Let vector field F =(x,y,z²) across the surface S which is an upper hemisphere x²+ y² + z² = 4 . 1)…

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Q: a) Is the vector field F = (sin z, y², x cos z) conservative? Explain your reasoning. b) Compute the…

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Q: A vector field is given as F = âr2r“ + a¿r²CoSo + âz5z. Verify Stokes theorem for this vector field…

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Q: Find the level surfaces of the scalar field U = ear, where the vectors are: a = (2, 2, 1) and r = =…

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Q: Verify F(x, y, z) = (2y + z)ê + (x – z)j + (y – z )k taken over a triangle cut from the Stokes'…

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Q: 26. Verify Stokes' Theorem for the vector field F = 2xyi + xj + (y+z)k and surface z = 4x² - y², z ≥…

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Q: 1: Let Let F = (yz², tan-1(xy), 1 + xy³). %3D a) Show that F is not conservative. b) Show that F…

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Q: r = 2 on the hemispherical surface defined by 0 ≤ Ɵ ≤ π/2 and 0 ≤ɸ ≤ 2π F = 100cosƟar Verify Stokes'…

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Q: Show that the regular surface S = {(x, y, z): x² + y² = 2² + 1, x > 0 or y > U} is orient: by…

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Q: Determine the flux of the vector field F(x, y, z) = (x, y, z) across the portion of the paraboloid z…

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Q: Verify Stoke's theorem for the vector field F = (2x - y) I– yz"J - y z K over the upper half surface…

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